$ A = \left[\begin{array}{rr}1 & 4 \\ 5 & 3 \\ -2 & 0\end{array}\right]$ $ F = \left[\begin{array}{rr}4 & 5 \\ 0 & -2\end{array}\right]$ What is $ A F$ ?
Because $ A$ has dimensions $(3\times2)$ and $ F$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(3\times2)$ $ A F = \left[\begin{array}{rr}{1} & {4} \\ {5} & {3} \\ \color{gray}{-2} & \color{gray}{0}\end{array}\right] \left[\begin{array}{rr}{4} & \color{#DF0030}{5} \\ {0} & \color{#DF0030}{-2}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ A$ , with the corresponding elements in column $j$ of the second matrix, $ F$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ A$ with the first element in ${\text{column }1}$ of $ F$ , then multiply the second element in ${\text{row }1}$ of $ A$ with the second element in ${\text{column }1}$ of $ F$ , and so on. Add the products together. $ \left[\begin{array}{rr}{1}\cdot{4}+{4}\cdot{0} & ? \\ ? & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ A$ with the corresponding elements in ${\text{column }1}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{1}\cdot{4}+{4}\cdot{0} & ? \\ {5}\cdot{4}+{3}\cdot{0} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ A$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{1}\cdot{4}+{4}\cdot{0} & {1}\cdot\color{#DF0030}{5}+{4}\cdot\color{#DF0030}{-2} \\ {5}\cdot{4}+{3}\cdot{0} & ? \\ ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{1}\cdot{4}+{4}\cdot{0} & {1}\cdot\color{#DF0030}{5}+{4}\cdot\color{#DF0030}{-2} \\ {5}\cdot{4}+{3}\cdot{0} & {5}\cdot\color{#DF0030}{5}+{3}\cdot\color{#DF0030}{-2} \\ \color{gray}{-2}\cdot{4}+\color{gray}{0}\cdot{0} & \color{gray}{-2}\cdot\color{#DF0030}{5}+\color{gray}{0}\cdot\color{#DF0030}{-2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}4 & -3 \\ 20 & 19 \\ -8 & -10\end{array}\right] $